Corrected exercises: power, exponential, product and quotient rules (MATH 140)

By the sum and constant multiple rules, each term is differentiated separately: $(4x^5)' = 4 \cdot 5x^4 = 20x^4$, $(-3x^2)' = -6x$, $(7x)' = 7$, and $(\pi^3)' = 0$ because $\pi^3 \approx 31$ is a NUMBER, not a power of $x$. So $f'(x) = 20x^4 - 6x + 7$. Writing $3\pi^2$ for the derivative of $\pi^3$ applies the power rule to a constant: the variable is $x$, and $\pi^3$ does not contain it. That single slip costs a mark on every paper where it appears.

Rewrite first: $\sqrt[3]{x} = x^{1/3}$, $\frac{1}{\sqrt x} = x^{-1/2}$ and $\frac{1}{x^2} = x^{-2}$, so $g(x) = 6x^{1/3} - 8x^{-1/2} + 5x^{-2}$. The power rule then gives $g'(x) = 6 \cdot \frac{1}{3}x^{-2/3} - 8 \cdot \left(-\frac{1}{2}\right)x^{-3/2} + 5 \cdot (-2)x^{-3} = 2x^{-2/3} + 4x^{-3/2} - 10x^{-3}$. Back to radicals: $g'(x) = \frac{2}{\sqrt[3]{x^2}} + \frac{4}{x\sqrt x} - \frac{10}{x^3}$, since $x^{3/2} = x\sqrt x$. The trap is the sign of the middle term: the exponent $-\frac{1}{2}$ multiplies $-8$, and two minus signs make $+4$. Subtracting $1$ from a negative exponent also moves it AWAY from zero: $-\frac{1}{2} - 1 = -\frac{3}{2}$, not $-\frac{1}{2} + 1$.

The denominator is a single power of $x$, so divide term by term instead of using the quotient rule: $h(x) = \frac{x^2}{x^{1/2}} - \frac{3x}{x^{1/2}} + \frac{5}{x^{1/2}} = x^{3/2} - 3x^{1/2} + 5x^{-1/2}$. Then $h'(x) = \frac{3}{2}x^{1/2} - \frac{3}{2}x^{-1/2} - \frac{5}{2}x^{-3/2}$. Over the common denominator $2x^{3/2} = 2x\sqrt x$: $\frac{3}{2}x^{1/2} = \frac{3x^2}{2x^{3/2}}$ and $\frac{3}{2}x^{-1/2} = \frac{3x}{2x^{3/2}}$, so $h'(x) = \frac{3x^2 - 3x - 5}{2x\sqrt x}$. At $x = 4$: $\sqrt 4 = 2$, so $h'(4) = \frac{48 - 12 - 5}{2 \cdot 4 \cdot 2} = \frac{31}{16}$. The quotient rule would give the same answer after twice the algebra; rewriting is the method a marker expects when the denominator is a monomial.

The three terms of $m$ look alike and are three different objects. $x^e$ is a POWER of $x$ with the constant exponent $e$: $(x^e)' = ex^{e-1}$. $e^x$ is the exponential, constant base and variable exponent: $(e^x)' = e^x$, and the power rule does not apply to it. $e^e$ is a constant: its derivative is $0$. So $m'(x) = ex^{e-1} + e^x$. For $p$, expand the square, which removes any composition: $p(x) = 4x + 4\sqrt x + 1 = 4x + 4x^{1/2} + 1$, so $p'(x) = 4 + 2x^{-1/2} = 4 + \frac{2}{\sqrt x}$. Check at $x = 1$: $p$ goes from $p(1) = 9$ with slope $6$, and indeed $4 + 2 = 6$.

First error: the student differentiated the denominator $3x^2$ into $6x$ and kept the fraction bar, as if $\left(\frac{1}{u}\right)' = \frac{1}{u'}$, a rule that does not exist. Second error, as a consequence: the sign is lost, although $\frac{1}{3x^2}$ is decreasing for $x > 0$, so its derivative there must be NEGATIVE. The correct route rewrites first: $\frac{1}{3x^2} = \frac{1}{3}x^{-2}$, so the derivative is $\frac{1}{3} \cdot (-2)x^{-3} = -\frac{2}{3x^3}$. At $x = 1$ the student's answer is $\frac{1}{6}$ and the true slope is $-\frac{2}{3}$: one test value is enough to expose the rule.

Product rule with $f(x) = x^2 + 1$, $f'(x) = 2x$, and $g(x) = x^3 - 2x$, $g'(x) = 3x^2 - 2$: $F'(x) = 2x(x^3 - 2x) + (x^2 + 1)(3x^2 - 2) = 2x^4 - 4x^2 + 3x^4 - 2x^2 + 3x^2 - 2 = 5x^4 - 3x^2 - 2$. Check by expanding first: $F(x) = x^5 - 2x^3 + x^3 - 2x = x^5 - x^3 - 2x$, so $F'(x) = 5x^4 - 3x^2 - 2$, the same. On a product of two polynomials both routes are legal; expanding is the natural CHECK of the product rule, and the factor-by-factor answer $2x(3x^2 - 2)$ fails it at once, being of degree $3$ instead of $4$.

Quotient rule with $f(x) = 3x - 1$, $f'(x) = 3$, and $g(x) = x^2 + 2$, $g'(x) = 2x$: $G'(x) = \frac{3(x^2 + 2) - (3x - 1)(2x)}{(x^2 + 2)^2} = \frac{3x^2 + 6 - 6x^2 + 2x}{(x^2 + 2)^2} = \frac{-3x^2 + 2x + 6}{(x^2 + 2)^2}$. Then $G'(0) = \frac{6}{4} = \frac{3}{2}$. The parentheses around $(3x - 1)(2x)$ carry the minus sign to BOTH terms: dropping them gives $-6x^2 - 2x$ and a wrong numerator. Writing $fg' - f'g$ in the numerator flips the sign of the whole answer; the order follows the rule's reading, derivative of the TOP first.

Product rule with $f(x) = x^2$ and $g(x) = e^x$, whose derivative is $e^x$ itself: $H'(x) = 2x \cdot e^x + x^2 \cdot e^x = (x^2 + 2x)e^x = x(x + 2)e^x$. Since $e^x > 0$ for every $x$, the product is zero exactly when $x(x + 2) = 0$: $x = 0$ or $x = -2$. Factoring $e^x$ out is not cosmetic: it is the step that turns an equation with an exponential into a polynomial equation, and the fact $e^x \ne 0$ must be SAID to justify dividing by it.

Quotient rule with $f(x) = 1 + e^x$, $f'(x) = e^x$, and $g(x) = 1 - e^x$, $g'(x) = -e^x$: $K'(x) = \frac{e^x(1 - e^x) - (1 + e^x)(-e^x)}{(1 - e^x)^2}$. The numerator is $e^x - e^xe^x + e^x + e^xe^x = 2e^x$, the two products $e^xe^x$ cancelling. So $K'(x) = \frac{2e^x}{(1 - e^x)^2}$, positive wherever it is defined. The double minus in $-(1 + e^x)(-e^x)$ is where most papers lose the answer: write the product $fg'$ in full, with its own sign, before expanding.

At $x = 1$, the true derivative from c) is $H'(1) = 1 \cdot 3 \cdot e = 3e$, while $2xe^x$ gives $2e$: the student multiplied the derivatives of the two factors and forgot the term $x^2e^x$. For the quotient, the true derivative is $\frac{1 \cdot (x + 1) - x \cdot 1}{(x + 1)^2} = \frac{1}{(x + 1)^2}$, which equals $\frac{1}{4}$ at $x = 1$, not $1$. Choosing the test value matters: at $x = 0$ both wrong answers happen to agree with the right ones ($0$ and $1$), so a single lucky value proves nothing. A test value can refute a rule, never confirm it.

Rewriting: $f(x) = \frac{x^3}{x^2} - \frac{4x}{x^2} + \frac{1}{x^2} = x - 4x^{-1} + x^{-2}$, so $f'(x) = 1 + 4x^{-2} - 2x^{-3} = 1 + \frac{4}{x^2} - \frac{2}{x^3}$. Quotient rule, with $(x^3 - 4x + 1)' = 3x^2 - 4$ and $(x^2)' = 2x$: $f'(x) = \frac{(3x^2 - 4)x^2 - (x^3 - 4x + 1)(2x)}{x^4} = \frac{3x^4 - 4x^2 - 2x^4 + 8x^2 - 2x}{x^4} = \frac{x^4 + 4x^2 - 2x}{x^4} = \frac{x^3 + 4x - 2}{x^3}$ after dividing by $x$. Over the denominator $x^3$, the first answer is also $\frac{x^3 + 4x - 2}{x^3}$: they agree. The rewriting took one line, the quotient rule four, with a denominator $x^4$ to simplify at the end.

Reciprocal rule with $c = 5$ and $g(x) = x^2 + 4$: $q'(x) = -\frac{5 \cdot 2x}{(x^2 + 4)^2} = -\frac{10x}{(x^2 + 4)^2}$, and $q'(1) = -\frac{10}{25} = -\frac{2}{5}$. The quotient rule gives the same, since the $f'g$ term vanishes: $\frac{0 \cdot (x^2 + 4) - 5 \cdot 2x}{(x^2 + 4)^2}$. The answer $\frac{0}{2x} = 0$ would mean that $q$ has a horizontal tangent everywhere, so that $q$ is constant; yet $q(0) = \frac{5}{4}$ and $q(1) = 1$. A constant NUMERATOR does not make the quotient constant.

For $x \ne 3$, $x^2 - 9 = (x - 3)(x + 3)$ and the factor $x - 3$ cancels: $g(x) = x + 3$, so $g'(x) = 1$. This is exactly the figure, a line of slope $1$. But $g$ is not defined at $x = 3$, which the open circle at $(3, 6)$ shows, so $g'(3)$ does not exist: a derivative at a point needs the function AT that point. Check with the quotient rule: $\frac{2x(x - 3) - (x^2 - 9)}{(x - 3)^2} = \frac{x^2 - 6x + 9}{(x - 3)^2} = 1$ for $x \ne 3$. Simplifying is allowed and fast, as long as the answer keeps the condition $x \ne 3$ that the simplification hides.

By the laws of exponents, $e^{x+2} = e^2e^x$, a constant times $e^x$: its derivative is $e^2e^x = e^{x+2}$. Next, $e^{-x} = \frac{1}{e^x}$, so by the reciprocal rule $(e^{-x})' = -\frac{e^x}{(e^x)^2} = -\frac{1}{e^x} = -e^{-x}$. Finally $e^{2x} = e^xe^x$, and the product rule gives $e^xe^x + e^xe^x = 2e^{2x}$. The student's claim is refuted by one look at the graph of $e^{-x}$: it DECREASES on the whole real line, so its tangent lines all have negative slopes, and a derivative equal to $e^{-x} > 0$ is impossible. The rule $(e^x)' = e^x$ is about the exponent $x$ exactly; any other exponent must be rewritten first here, or handled by the chain rule later.

Divide first: $x^2 + 1 = (x^2 - 1) + 2$, so $w(x) = 1 + \frac{2}{x^2 - 1}$, for $x \ne \pm 1$. The constant $1$ has derivative $0$ and the reciprocal rule gives $w'(x) = -\frac{2 \cdot 2x}{(x^2 - 1)^2} = -\frac{4x}{(x^2 - 1)^2}$. Check with the quotient rule: $\frac{2x(x^2 - 1) - (x^2 + 1)(2x)}{(x^2 - 1)^2} = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2}$. Same answer. When numerator and denominator have the same degree, one division turns the quotient into a constant plus a simpler fraction.

By the power, sum and constant multiple rules, $f'(x) = 3x^2 - 6x - 9$. At $a = 1$: $f(1) = 1 - 3 - 9 + 5 = -6$ and $f'(1) = 3 - 6 - 9 = -12$. The tangent is $y = -6 - 12(x - 1) = -12x + 6$. Check: at $x = 1$ the line gives $-12 + 6 = -6 = f(1)$, so it does pass through the point. On the figure the curve is indeed going down steeply at $x = 1$. The form $y = -6 + (3x^2 - 6x - 9)(x - 1)$, with $f'(x)$ left unevaluated, is a CUBIC, not a line, and earns no mark: substitute $a$ into $f'$ before writing the equation.

The tangent has slope $-12 \ne 0$, so the normal has slope $-\frac{1}{-12} = \frac{1}{12}$ and passes through $(1, -6)$: $y = -6 + \frac{1}{12}(x - 1) = \frac{1}{12}x - \frac{73}{12}$. The check is the product of the slopes: $(-12) \cdot \frac{1}{12} = -1$, the condition for two non-vertical lines to be perpendicular. Answering with slope $12$ (the opposite, forgetting the reciprocal) or $-\frac{1}{12}$ (the reciprocal, forgetting the sign) are the two classic errors, and each costs the question.

A horizontal tangent means slope $0$: $3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) = 0$, so $x = -1$ or $x = 3$. Then $f(-1) = -1 - 3 + 9 + 5 = 10$ and $f(3) = 27 - 27 - 27 + 5 = -22$: the points are $(-1, 10)$ and $(3, -22)$, the top of the hump and the bottom of the valley on the figure. At each of them the tangent is horizontal, $y = 10$ and $y = -22$, and the normal is VERTICAL: $x = -1$ and $x = 3$. The formula $-\frac{1}{f'(a)}$ breaks down here, and writing no normal exists is wrong: a vertical line is a line, it simply has no slope.

Parallel lines have equal slopes, so solve $f'(x) = 15$: $3x^2 - 6x - 9 = 15$, $3x^2 - 6x - 24 = 0$, $x^2 - 2x - 8 = 0$, $(x - 4)(x + 2) = 0$. So $x = 4$ or $x = -2$, with $f(4) = 64 - 48 - 36 + 5 = -15$ and $f(-2) = -8 - 12 + 18 + 5 = 3$. The points are $(4, -15)$ and $(-2, 3)$. The question asks for POINTS, so both coordinates are expected; stopping at $x = 4$ and $x = -2$ loses part of the marks.

Complete the square: $3x^2 - 6x - 9 = 3(x^2 - 2x + 1) - 3 - 9 = 3(x - 1)^2 - 12$. Since $(x - 1)^2 \ge 0$, every slope satisfies $f'(x) \ge -12$, with equality only at $x = 1$. The equation $f'(x) = -13$ would need $3(x - 1)^2 = -1$, impossible: no tangent has slope $-13$. The smallest slope of a tangent is $-12$, reached at $x = 1$, which is why the tangent of a) looked like the steepest descent on the figure. Every question about the slopes of a curve is a question about the FUNCTION $f'$, here a parabola that opens upward.